Prove that the degree of the minimal polynomial over $\mathbb{Q}$ is a power of $2$ 
$\textbf{Question.}$ Let $n$  a positive integer and let be $K = \mathbb{Q}[a_1, \cdots, a_n]$ where $a_i$ is such that $a_i^2 \in \mathbb{Q}$ for each $i = 1, \cdots, n$. Given $b \in K$, prove that the degree of the minimal polynomial $p_b(x)$ of $b$ over $\mathbb{Q}$ is a power of $2$.

I would like a hint to start to solve this problem, because I don't have ideas how to start it.
 A: Consider the minimal polynomial of $a_{i+1}$ over $\Bbb{Q}[a_1,...,a_i]$.  It divides $x^2-a_{i+1}^2\in\Bbb{Q}[x]$ and so must have degree $1$ or $2$.  But this means the degree of $\Bbb{Q}[a_1,...,a_i,a_{i+1}]$ over $\Bbb{Q}[a_1,...,a_i]$ is $1$ or $2$.  Thus, if the degree of $\Bbb{Q}[a_1,...,a_i]$ over $\Bbb{Q}$ is a power of $2$, say $2^j$, then the degree of $\Bbb{Q}[a_1,...,a_i,a_{i+1}]$ over $\Bbb{Q}$ is the product of the degree of $\Bbb{Q}[a_1,...,a_i,a_{i+1}]$ over $\Bbb{Q}[a_1,...,a_i]$ and that of $\Bbb{Q}[a_1,...,a_i]$ over $\Bbb{Q}$, and is thus $2^j$ or $2^{j+1}$, in either case a power of $2$. By induction, then, the degree of $\Bbb{Q}[a_1,...,a_n]$ over $\Bbb{Q}$is a power of $2$.
A: Firstly, I would like to thanks Jyrki Lahtonen by the hints given in the comments of my OP and which enabled me to answer my own question.
Following the hints of Jyrki Lahtonen, I start by induction proving that $[ \mathbb{Q}[a_1, \cdots, a_n] : \mathbb{Q} ] = 2^m$ for some $m \in \mathbb{N}$.
Induction Base ($n = 1$):
Let be $a_1$ such that $a_1^2 \in \mathbb{Q}$ and $K := \mathbb{Q}(a_1)$. Since $a_1^2 \in \mathbb{Q}$, we know that $f(x) := x^2 - a_1^2 \in \mathbb{Q}[X]$ and $a_1$ is a root of $f$, then the minimal polynomial $p_{a_1}(x)$ divides f(x), therefore $p_{a_1}(x)$ has degree $1$ or $2$, i.e.,
$$[\mathbb{Q}(a_1) : \mathbb{Q}] = 2^m, m = 0 \ \text{or} \ 1$$
Supposing that $[\mathbb{Q} (a_1, \cdots, a_n) : \mathbb{Q}] = 2^m$ for some $m \in \mathbb{N}$ and $a_i^2 \in \mathbb{Q}$ for each $i = 1,\cdots, n + 1$ we observe that
$$[\mathbb{Q}(a_1, \cdots, a_{n+1}) : \mathbb{Q}(a_1, \cdots, a_n)] = 2^p, p = 0 \ \text{or} \ 1,$$
by the same argument used in base step, just redefining $f(x) := x^2 - a_{n+1}^2$. By the tower of extensions and induction hypothesis, we know that
$$[\mathbb{Q}(a_1, \cdots, a_{n+1}) : \mathbb{Q}] = [\mathbb{Q}(a_1, \cdots, a_{n+1}) : \mathbb{Q}(a_1, \cdots, a_n)] \cdot [\mathbb{Q}(a_1, \cdots, a_n) : \mathbb{Q}] = 2^p \cdot 2^m = 2^{p + m},$$
then the induction proof is complete.
Following another hint of Jyrki Lahtonen, we observe that $\mathbb{Q}(b) \subset K$, because $b \in K$ by the statement of the question, then $\mathbb{Q} \subset \mathbb{Q}(b) \subset K$ and
$$[K : \mathbb{Q}] = [K : \mathbb{Q}(b)] \cdot [\mathbb{Q}(b) : \mathbb{Q}]$$
by the tower of the extensions, but observe that $x - b$ is a minimal polynomial over $\mathbb{Q}(b)$ because $b \in K$, this imply that $[K : \mathbb{Q}(b)] = \text{degree(x - b)} = 1$, so
$$[\mathbb{Q}(b) : \mathbb{Q}] = [K : \mathbb{Q}] = 2^m, \ m \in \mathbb{N}$$
by the statement proved by induction, but this means that the degree of a minimal polynomial of $b$ over $\mathbb{Q}$ is $2^m$, i.e., is a power of $2$. $\square$
